Fundamentals of Structural Analysis Episode 1 Part 3 pps

Before we proceed to find the force unknowns by the method of joint, we must be sure that all the force unknowns can be determined by the static equilibrium conditions alone, because tha

Solution. We shall give a detailed step-by-step solution.

(1) Identify all force unknowns The very first step in any force method of analysis is to

identify all force unknowns This is achieved by examining the reaction forces and member forces The reaction forces are exposed in a FBD of the whole structure as shown

Free-Body diagram of the three-bar-truss to expose the reaction forces.

Note that in the above figure the subscripts of the reaction forces indicate the direction (first subscript) and the location of the reactions (second subscript) The three reaction

forces are Rx1, R y1 and R y3 The member forces are F1, F2, and F3

(2) Examine the static determinacy of the structure Before we proceed to find the

force unknowns by the method of joint, we must be sure that all the force unknowns can

be determined by the static equilibrium conditions alone, because that is the essence of the method of joint, namely using joint equilibrium equations to find force unknowns

Denote the number of all member force unknowns as M and the number of reaction forces as R, the total number of force unknowns is M+R In the present example, M=3,

R=3 and M+R =6 This number is to be compared to the number of equilibrium equations

available

There are three nodes in the truss We can write two equilibrium equations at each node

of a plane truss:

Thus the total number of equilibrium equations available is 2N, where N is the number of nodes in a truss In the present example, N=3 and 2N=6 Thus, the number of available

static equilibrium equations matches the total number of force unknowns exactly,

M+R=2N The problem posed in the present example is statically determinate We can

reach the same conclusion if we note that the truss is a simple truss

3

2

3 1

1.0 kN 0.5 kN

R x1

(3) Solve for force unknowns The most obvious next step is to write up the six nodal

equilibrium equations and solve for the six unknown forces simultaneously That would require the use of a computer For the present example, and many other cases, an

experienced structure engineer can solve a problem by hand calculation faster than using

a computer This hand-calculation process gives insight to the force flow from externally applied load, through members, and to the supports This is the process that is presented herein

(a) Find all reactions Although not necessary, finding all reaction forces from

the FBD of the whole structure first is often the fastest way of solving a plane truss problem

Free-Body-Diagram for finding reactions.

The three reaction forces can be solved one at a time by applying the three equilibrium equations one by one:

Σ M1 = 0 R y3(6) –(1.0)(3) –(0.5)(4) = 0 R y3 = 0.83 kN

(b) Find member forces The member forces are solved by applying nodal equilibrium equations joint by joint The selection of the sequence by which each joint is

3

2

3 1

1.0 kN 0.5 kN

R x1

4m

Joint 3.

Σ Fy = 0, F2(4/5)+0.83=0, F2= –1.04 kN

Σ Fx = 0, –F2(3/5) – F3=0, F3= 0.62 kN

Joint 1.

Σ Fy = 0, F1(4/5)+0.17=0, F1= –0.21 kN

Note that only one equation from the FBD of joint 1 is needed to find the remaining

unknown of F1 The second equilibrium equation is identically satisfied The two

equilibrium equations from the FBD of joint 2 would also be identically satisfied These three “unused” equations can serve as a “check” for the accuracy of the computation We need not use these three joint equations because we have already used three equations from the equilibrium of the whole structure at the beginning of the solution process This fact also points to an important point: There are no more than six independent

equilibrium equations Any additional equations are not “independent” from the six equations we just used because they can be derived from the linear combination of the six equations Any six “independent” equations are equally valid The selection of which six equations to use is a matter of preference and we always select those equations that give

us the easiest way of getting the answer to the unknown forces as we just did

Example 2 Find all reaction and member forces for the loaded truss shown.

Another truss example problem for the method of joint.

Solution. A slightly different solution strategy is followed in this example

F2

3

F3

0.83 kN

1 0.5 kN

0.17 kN

F1

0.62 kN

x y

1

2

3m

2m

3

2m

3

6 kN

1.5 m

4

3

4 5

3 4 5

4

5

(1) Identify all force unknowns The FBD of the whole structure shows there are four

reactions Adding the six member forces, we have M=6, R=4 and M+R=10, a total of

ten force unknowns

FBD of the whole truss.

(2) Examine the static determinacy of the structure There are five nodes, N=5 Thus

M+R=2N=10 This is a statically determinate problem.

(3) Solve for force unknowns This is a problem for which there is no advantage in

solving for the reactions first The FBD of the whole structure will give us three

equations of equilibrium while we have four reaction unknowns Thus, we cannot solve for the four reactions with the equations from the FBD of the whole structure alone On the other hand, if we go from joint to joint in the following order, 3, 2, 4, 1, and 5, we will be able to solve for member forces one node at a time and eventually getting to the reactions

Joint 3.

Σ Fy = 0, F5(3/5)+ F6(3/5)= –6,

Σ Fx = 0, –F5(4/5) + F6(4/5) =0

F5 = – 5 kN, F6= – 5 kN

In this case, solving the two equations simultaneously is inevitable

2

3

6 kN

4

R y1

R x1

1

3

4

5 R x5

R y5

F5

3 4 5

6 kN 3

F6

3 4

5

Joint 4.

Σ Fx = 0, F6(4/5) + F3(4/5) =0, F3 = 5 kN

Σ Fy = 0, F6(3/5) – F3(3/5) –F2= 0,

F2= –6 kN

Joint 1.

Σ Fx = 0, R x1+ F3(4/5) =0, R x1 = –4 kN.

Σ Fy = 0, R y1+F3(3/5) +F1= 0,

R y1= 3 kN.

Joint 5.

Σ Fx = 0, R x5 – F4(4/5) =0, R x5 = 4 kN

Σ Fy = 0, R y5 +F4(3/5) +F2= 0,

R y5= 3 kN.

Note in both example problems, we always assume the member forces to be in tension This results in FBDs that have member forces pointing away from the joints This is simply an easy way to assign force directions It is highly recommended because it avoids unnecessary confusion that often leads to mistakes

Example 3 Find the member forces in bars 4, 5, 6, and 7 of the loaded Fink truss shown.

Fink truss to be solved by the method of joint.

4

F6

F3 F

2

4 3 4 5

1

F3

F1

R y1

4 5

F4 F2

R y5

5 R x5

4

10 kN 3@2m=6m

5

6

7

7 8

10

9

4

Solution We shall illustrate a special feature of the method of joint.

(1) Identify all force unknowns The FBD of the whole structure would have shown

that there are three reactions Adding the eleven member forces, we have M=11, R=3 and

M+R=14, a total of 14 force unknowns.

(2) Examine the static determinacy of the structure There are seven nodes, N=7.

Thus M+R=2N=14 This is a statically determinate problem.

(3) Solve for force unknowns Normally Fink trusses are used to take roof loading on

the upper chord nodes We deliberately apply a single load at a lower chord node in order to make a point about a special feature of the method of joint We start by

concentrating on Joint 5

Joint 5.

Σ Fy = 0, F4 = 0

Σ Fx = 0, –F8 + F9 =0

F8= F9

In this case, it is advantageous to line up the coordinate system with the local geometry at

the node F4 is found to be zero because it is the only force in that direction The pair of

forces in the x-direction must be equal and opposite because they are co-linear.

Joint 2.

F1 = F2

Joint 7.

5

x y

F4

F8

F9

2

F5

F4

F2

F1

7

F10

F11

F7

Joint 3.

F7 = 0, from equilibrium of Joint 7

Σ Fy = 0, F6 (2/2.23) = 10 F6 = 11.15 kN

That completes the solution for F4,F5,F6 and F7

Thus, with the exception of member 6, all the web members are zero-force members for this particular loading case For purpose of analysis under the given load the Fink truss is equivalent to the truss shown below

Equivalent truss to the Fink truss for the given load.

This brings up the interesting feature of the method of joint: we can identify zero-force members easily This feature is further illustrated in the next example

Example 4 Identify zero-force members and equal-force members in the loaded trusses

shown

An example of zero-force members and equal force members.

Solution. The equilibrium of forces at joint C leads to FCG=0 and FBC=FCD Once we know F CG = 0, it follows F BG =0 and then F BF = 0, based on the equilibrium of forces at

node G and node B, respectively The equilibrium of forces at joint F leads to F AF =P and

F EF=FFG

3

10 kN

F3

F2

1

2 2.23

P

D

E

F

G

P

F CG

F CD

F BC

C

F FG

F EF

F AF

10 kN

4 m

2m

2 m

We can identify:

(1) zero force members At each joint, all the forces are concurrent forces If all the forces are co-linear except one then the lone exception must be zero

(2) equal force members If two forces at a joint are co-linear and all other forces at the joint are also co-linear in another direction, then the two forces must be equal

An equivalent truss.

For practical purposes, the original truss problem is equivalent to the truss problem shown above for the given loading case

P

A

D E

Problem 1 Use the method of joint to find all reaction and member forces in the trusses

shown

Problem 1.

3m

4m

3 kN

3m

4m

3m

8 kN

3m

4m

3m

8 kN

3 kN

3m

4m

4 kN

3m

4m

4 kN

3m

2m

6 kN

2m 1.5m

1.2 m

1.6 m 0.9 m

2 m

2 m

1.2 m

0.9 m 0.7 m

4kN

0.9 m

5kN 1.2 m

1m 1m

1 kN

2 m

2 m

1m 1m

2 kN

2 m

2 m

1m 1m

2 kN

1 kN

Example 5 Find member forces in bars in the 3rd panel from the left of the truss shown.

An example problem for the method of section.

Solution. We shall solve this problem by the method of section with the following

procedures

(1) Name all joints We can refer to each joint by a symbol and each member by the two

end joints as shown in the figure below We also define an x-y coordinate system as shown We need to find FIJ, FCJ, and FCD The truss is stable and determinate.

(2) Find reactions We have to look at the FBD of the whole truss.

The FBD to find the reactions.

ΣMA=0, (12) (30) – (18) RGV = 0, R GV = 20 kN.

4 m

6@3 m = 18 m 30 kN

4 m

6@3 m = 18 m 30 kN

A

R AH

x

y

forces to deal with We always assume the member forces are tensile We have already

obtained RAV = 10 kN.

FBD exposing a section through the third panel from left.

ΣMC=0, (4) FIJ + (6) RAV = 0, F IJ = – 1.5 RAV = –l15 kN.

ΣMJ=0, – (4) FCD + (9) RAV = 0, F CD = 2.25 RAV = 22.5 kN.

ΣFy=0, (0.8) FCJ + RAV = 0, F CJ = –1.25 RAV = –12.5 kN.

Note that we choose the moment center at C and J, respectively, because in each case the

resulting equation has only one unknown and therefore can be solved easily

To illustrate the effect of taking a different FBD, let us choose the right part of the cut as

the FBD Note that we already know R GV = 20 kN

Alternative FBD exposing the member forces of the 3 rd panel.

By taking the right portion as the FBD we include the applied 30 kN force in the FBD and it will show up in all equilibrium equations

ΣMC=0, – (4) FIJ + (6) (30) – (12) RGV = 0, FIJ = – 3 RGV + 45 = –15 kN ΣMJ=0, (4) FCD + (3) (30) – (9) RGV = 0, FCD = 2.25 RGV – 22.5= 22.5 kN

A

R AV

x y

F CJ

F CD

4 m

3@ 3m=9 m

4@ 3m=12 m

x

30 kN

C

R GV

F IJ

F CJ

F CD

ΣFy=0, – (0.8) FCJ – 30 + R GV = 0, F CJ = –37.5 +1.25 RGV = –12.5 kN.

Example 6 Find member forces in bars in the 2nd panel from the left of the truss shown

Another example problem for the method of section.

Solution. The inclined chord geometry will cause complications in computation, but the process is the same as that of the last example

(1) Find reactions This is a simple truss, stable and determinate.

FBD for reaction forces.

ΣMA=0 – (16) REV+(4)3+(8)6+(12)9=0, R EV =10.5 kN.

A

B

C

F

G

H

4 @ 4m=16 m

2m 3m

A

B

C

F

G

H

4 @ 4m=16 m

2m 3m

R EV

R AV

R AH

x y

FBD for the 2nd panel member forces.

In order to find FBC we want to find a moment center that is the intersection of the two other unknowns The intersection point of F FG and F FC is point F.

Similarly, we take moment about point C so that the only unknown force in the ensuing equilibrium equation would be F FG In writing the moment equilibrium equation, we

utilize the fact that FFG can be transmitted to point K and the horizontal component of

F FG at K has no contribution to the equilibrium equation while the vertical component is (2/4.47) F FG =0.447 FFG has, as shown in the left figure below.

ΣMC=0 (10) 0.447FFG+(8)7.5– (4)3=0, F FG = –10.74 kN.

Two FBDs to find F FG

Alternatively we can transmit F FG to point G, and use the horizontal component

(4/4.47) FFG=0.894 FFG in the moment equation, as shown in the right figure above.

ΣMC=0 (5) 0.894F FG + (8)7.5– (4)3=0, F FG = –10.72 kN

A

3 kN

B F

7.5 kN

FFG

FFC

FBC

3m

C

A

3 kN

B F

7.5 kN

K

F FG

FFC

F BC

3m

0.447F FG

2m

C

F FG

4 2 4.47

A

3 kN

B F

7.5 kN

F FG

F FC

F BC

3m

G

2m

F FG

0.894F FG

C

To find FFC we need to go out of the region of the truss to find the moment center (K) as the shown in the left figure above, and use the vertical component of the transmitted F FC

at point C.

ΣMK=0 (10) 0.6F FC – (2)7.5+(6)3=0, F FC = -0.50 kN

Note that all these additional efforts are caused by the inclined upper chord of the truss

Example 7 Find the force in the top and bottom chord members of the third panel from

the left of the K-truss shown.

K-truss example.

Solution. The K-truss is a simple truss that requires a special cut for the solution of top

and bottom chord member forces as we shall see shortly It is stable and determinate

(1) Find reactions Since the truss and the loading are symmetric, the reactions at both

supports are easily found to be 8 kN upward and there is no horizontal reaction at the left support

(2) Establish FBD The special cut is shown by the dotted line below.

6@3m=18m

4m 4m

16 kN

This particular cut separates the truss into two parts We shall use the left part as the FBD

FBD for top and bottom chord member forces.

Although there are four forces at the cut, two of them are on the same line When

moment center is selected at either node E or node G, these two forces will not appear in

the equilibrium equation, leaving only one unknown in each equation

ΣME=0, (6) 8 – (8) F GJ =0 F GJ = 6 kN

ΣMG=0, (6) 8 + (8) F EH =0 F EH = −6 kN

Alternatively, we may choose the right part as the FBD The same results will follow but the computation is slightly more involved

Alternative FBD for top and bottom chord member forces.

ΣME=0, (12) 8 – (3) 16 – (8) FGJ =0 F GJ = 6 kN.

ΣMG=0, (12) 8 – (3) 16 + (8) F EH =0 F EH = −6 kN

B

D

8 kN

C A

F EH

G

F GJ

J

8m

6m

J

H E

G

F EH

F GJ

8m

Example 8 Find the force in the inclined web members of the third panel from the left

of the K-truss shown

K-Truss example – inclined web members.

Solution A different cut is needed to expose the web member forces.

(1) Establish FBD To expose the force in the inclined web members, we may make a

cut through the third panel

A cut to expose web member forces.

This cut exposes four forces, the top and bottom member forces which are known and the

two inclined web member forces, FFH and FFJ, which are unknown.

6@3m=18m

4m 4m

16 kN

16 kN

F

J H

H F